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$$\sum_{k=1}^nk\binom nk$$ I have to write this as $n2^{n-1}$, and nothing I tried has worked so far. How should I approach this? Thank you

Alexander
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  • "and nothing I tried has worked so far" Such as? – Did Aug 19 '17 at 08:03
  • First I tried writing everything w.r.t. 2^n, like: n2^n-(n-1)(Combinations of n taken 1 at a time)-(n-2)*(C n,2)-...-(C n,n-1) but I kind of stopped here as it didn;t help much. Then I tried converting everything to Arrangements, but that's definitely not the good approach as the result is obtained from the sum of Combinations... – Alexander Aug 19 '17 at 08:08
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    "I kind of stopped here as it didn;t help much" Huh? Works perfectly, actually... You rightfully noted that the sum $S_n$ to be computed is $$S_n=\sum_{k=1}^nk{n\choose k}=\sum_{k=1}^n(n-(n-k)){n\choose k}=n\sum_{k=1}^n{n\choose k}-\sum_{k=1}^n(n-k){n\choose k}$$ thus, using the change of variable $i=n-k$, $$S_n=n(2^n-1)-\sum_{i=0}^{n-1}i{n\choose n-i}$$ and, since $${n\choose n-i}={n\choose i}$$ you get $$S_n=n(2^n-1)-\sum_{i=1}^{n-1}i{n\choose i}=n(2^n-1)-(S_n-n)$$ and you are done. – Did Aug 19 '17 at 08:51

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Hint

Consider $$\sum_{k=1}^nk\binom nk x^{k-1}$$ which the derivative of $$\sum_{k=1}^n\binom nk x^{k}=-1+\sum_{k=0}^n\binom nk x^{k}$$

When done, set $x=1$

Claude Leibovici
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